Backspace String Compare — LeetCode 844 Python Solution
EasyStackTwo PointersStringSimulation
- Problem
- #844
- Pattern
- Stack
- Reading time
- 5 min
- Source
- leetcode.com
The problem
Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.
Example
- Input
- s = "ab#c", t = "ad#c"
- Output
- true
- Explanation
- Both s and t become "ac".
Python solution
Python
class Solution:
def backspaceCompare(self, s: str, t: str) -> bool:
i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
while i >= 0 or j >= 0:
while i >= 0:
if s[i] == '#':
skip1 += 1
i -= 1
elif skip1:
skip1 -= 1
i -= 1
else:
break
while j >= 0:
if t[j] == '#':
skip2 += 1
j -= 1
elif skip2:
skip2 -= 1
j -= 1
else:
break
if i >= 0 and j >= 0:
if s[i] != t[j]:
return False
elif i >= 0 or j >= 0:
return False
i, j = i - 1, j - 1
return TrueComplexity
| Measure | Complexity |
|---|---|
| Time | O(n) (after optional sort O(n log n)) |
| Space | O(1) auxiliary |
Pattern: Stack
When the most recent unresolved thing is the one that matters, use a stack. LeetCode 844. Backspace String Compare is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Stack.
The stack guide has the Python template for the pattern and the 194 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 844. Backspace String Compare?
- LeetCode 844. Backspace String Compare is rated Easy on LeetCode.
- What is the time complexity of LeetCode 844. Backspace String Compare?
- The Python solution on this page runs in O(n) (after optional sort O(n log n)).
- What is the space complexity of LeetCode 844. Backspace String Compare?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 844. Backspace String Compare cover?
- LeetCode 844. Backspace String Compare is tagged Stack, Two Pointers, String and Simulation on LeetCode.